Solve 1/x+3+1/x=13/40

$$\frac { 1 } { x + 3 } + \frac { 1 } { x } = \frac { 13 } { 40 }$$

Solve for x

$$x = -\frac{24}{13} = -1\frac{11}{13} \approx -1.846153846$$

$$x=5$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -3,0 since division by zero is not defined. Multiply both sides of the equation by 40x\left(x+3\right), the least common multiple of x+3,x,40.

$$40x+40x+120=13x\left(x+3\right)$$

Combine 40x and 40x to get 80x.

$$80x+120=13x\left(x+3\right)$$

Use the distributive property to multiply 13x by x+3.

$$80x+120=13x^{2}+39x$$

Subtract 13x^{2} from both sides.

$$80x+120-13x^{2}=39x$$

Subtract 39x from both sides.

$$80x+120-13x^{2}-39x=0$$

Combine 80x and -39x to get 41x.

$$41x+120-13x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-13x^{2}+41x+120=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -13x^{2}+ax+bx+120. To find a and b, set up a system to be solved.

$$a+b=41 ab=-13\times 120=-1560$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -1560.

$$-1,1560 -2,780 -3,520 -4,390 -5,312 -6,260 -8,195 -10,156 -12,130 -13,120 -15,104 -20,78 -24,65 -26,60 -30,52 -39,40$$

Calculate the sum for each pair.

$$-1+1560=1559 -2+780=778 -3+520=517 -4+390=386 -5+312=307 -6+260=254 -8+195=187 -10+156=146 -12+130=118 -13+120=107 -15+104=89 -20+78=58 -24+65=41 -26+60=34 -30+52=22 -39+40=1$$

The solution is the pair that gives sum 41.

$$a=65 b=-24$$

Rewrite -13x^{2}+41x+120 as \left(-13x^{2}+65x\right)+\left(-24x+120\right).

$$\left(-13x^{2}+65x\right)+\left(-24x+120\right)$$

Factor out 13x in the first and 24 in the second group.

$$13x\left(-x+5\right)+24\left(-x+5\right)$$

Factor out common term -x+5 by using distributive property.

$$\left(-x+5\right)\left(13x+24\right)$$

To find equation solutions, solve -x+5=0 and 13x+24=0.

$$x=5 x=-\frac{24}{13}$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -3,0 since division by zero is not defined. Multiply both sides of the equation by 40x\left(x+3\right), the least common multiple of x+3,x,40.

$$40x+40x+120=13x\left(x+3\right)$$

Combine 40x and 40x to get 80x.

$$80x+120=13x\left(x+3\right)$$

Use the distributive property to multiply 13x by x+3.

$$80x+120=13x^{2}+39x$$

Subtract 13x^{2} from both sides.

$$80x+120-13x^{2}=39x$$

Subtract 39x from both sides.

$$80x+120-13x^{2}-39x=0$$

Combine 80x and -39x to get 41x.

$$41x+120-13x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-13x^{2}+41x+120=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -13 for a, 41 for b, and 120 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-41±\sqrt{41^{2}-4\left(-13\right)\times 120}}{2\left(-13\right)}$$

Square 41.

$$x=\frac{-41±\sqrt{1681-4\left(-13\right)\times 120}}{2\left(-13\right)}$$

Multiply -4 times -13.

$$x=\frac{-41±\sqrt{1681+52\times 120}}{2\left(-13\right)}$$

Multiply 52 times 120.

$$x=\frac{-41±\sqrt{1681+6240}}{2\left(-13\right)}$$

Add 1681 to 6240.

$$x=\frac{-41±\sqrt{7921}}{2\left(-13\right)}$$

Take the square root of 7921.

$$x=\frac{-41±89}{2\left(-13\right)}$$

Multiply 2 times -13.

$$x=\frac{-41±89}{-26}$$

Now solve the equation x=\frac{-41±89}{-26} when ± is plus. Add -41 to 89.

$$x=\frac{48}{-26}$$

Reduce the fraction \frac{48}{-26} to lowest terms by extracting and canceling out 2.

$$x=-\frac{24}{13}$$

Now solve the equation x=\frac{-41±89}{-26} when ± is minus. Subtract 89 from -41.

$$x=-\frac{130}{-26}$$

Divide -130 by -26.

$$x=5$$

The equation is now solved.

$$x=-\frac{24}{13} x=5$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -3,0 since division by zero is not defined. Multiply both sides of the equation by 40x\left(x+3\right), the least common multiple of x+3,x,40.

$$40x+40x+120=13x\left(x+3\right)$$

Combine 40x and 40x to get 80x.

$$80x+120=13x\left(x+3\right)$$

Use the distributive property to multiply 13x by x+3.

$$80x+120=13x^{2}+39x$$

Subtract 13x^{2} from both sides.

$$80x+120-13x^{2}=39x$$

Subtract 39x from both sides.

$$80x+120-13x^{2}-39x=0$$

Combine 80x and -39x to get 41x.

$$41x+120-13x^{2}=0$$

Subtract 120 from both sides. Anything subtracted from zero gives its negation.

$$41x-13x^{2}=-120$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-13x^{2}+41x=-120$$

Divide both sides by -13.

$$\frac{-13x^{2}+41x}{-13}=-\frac{120}{-13}$$

Dividing by -13 undoes the multiplication by -13.

$$x^{2}+\frac{41}{-13}x=-\frac{120}{-13}$$

Divide 41 by -13.

$$x^{2}-\frac{41}{13}x=-\frac{120}{-13}$$

Divide -120 by -13.

$$x^{2}-\frac{41}{13}x=\frac{120}{13}$$

Divide -\frac{41}{13}, the coefficient of the x term, by 2 to get -\frac{41}{26}. Then add the square of -\frac{41}{26} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{41}{13}x+\left(-\frac{41}{26}\right)^{2}=\frac{120}{13}+\left(-\frac{41}{26}\right)^{2}$$

Square -\frac{41}{26} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{41}{13}x+\frac{1681}{676}=\frac{120}{13}+\frac{1681}{676}$$

Add \frac{120}{13} to \frac{1681}{676} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{41}{13}x+\frac{1681}{676}=\frac{7921}{676}$$

Factor x^{2}-\frac{41}{13}x+\frac{1681}{676}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{41}{26}\right)^{2}=\frac{7921}{676}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{41}{26}\right)^{2}}=\sqrt{\frac{7921}{676}}$$

Simplify.

$$x-\frac{41}{26}=\frac{89}{26} x-\frac{41}{26}=-\frac{89}{26}$$

Add \frac{41}{26} to both sides of the equation.

$$x=5 x=-\frac{24}{13}$$